How do you factor #81x ^ { 2} - 144#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Narad T. Jan 27, 2017 The answer is #=9(3x+4)(3x-4)# Explanation: We use #a^2-b^2=(a+b)(a-b)# Therefore, #81x^2-144 =9(9x^2-16)# #=9(3x+4)(3x-4)# Answer link Related questions How do you factor special products of polynomials? How do you identify special products when factoring? How do you factor #x^3 -8#? What are the factors of #x^3y^6 – 64#? How do you know if #x^2 + 10x + 25# is a perfect square? How do you write #16x^2 – 48x + 36# as a perfect square trinomial? What is the difference of two squares method of factoring? How do you factor #16x^2-36# using the difference of squares? How do you factor #2x^4y^2-32#? How do you factor #x^2 - 27#? See all questions in Factor Polynomials Using Special Products Impact of this question 2819 views around the world You can reuse this answer Creative Commons License